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I started to learn stochastic processes this year. Only had two classes, but I already have some problem. We learned about Einstein's and Langevin's description of Brownian-motion and now I need to solve a task, related to Brownian-motion/rand-walk.
The task's text is the following (sorry, maybe it turned out as a very raw translation):

To understand the Perrin-experiment, consider the following variation of the 2D Brownian-motion: On a 2D lattice with lattice constant $l$ there is a moving particle. In every $\tau$ intervals, there is an equal probability that the particle jumps into any of the 4 neighbouring lattice points. These events are independent from each other. At $t = 0$, the particle is staying in the $(x_{0} = 0, y_{0} = 0)$ point.
Calculte the expected value of the displacement $\sqrt{\left< r_{t}^{2} \right>} = \sqrt{\left< x_{t}^{2} \right> + \left< y_{t}^{2} \right>}$, after $t = N \cdot \tau$.

I already solved the 1D brownian motion (and also rand-walk), using the Chapman-Kolmogorov-equation, then the Kramers-Moyal expansion to get the Fokker-Planck-equation:

$$ P \left( x, t + \tau \right) = \int_{- \infty}^{\infty} \Phi \left( \Delta \right) P \left( x - \Delta, t \right)\, d \Delta \quad \text{(Chapman-Kolmogorov)}$$

$$\vdots$$

$$ \frac{\partial P \left( x, t \right)}{\partial t} = \frac{\overline{\Delta^{2}}}{2 \tau} \frac{\partial^{2} P \left( x, t \right)}{\partial x^{2}} \quad \text{(Fokker-Planck)} $$

Then naturally said, that "oh wow, what a coincidence, that's totally some Gaussian, if the initial condition is $P \left( x, t=0 \right) = \delta \left( x \right)$" (so the start is from the center of the 1D coordinate system). And then I easily found $\left< x \right> = 0$ and $\left< x^{2} \right> = 2Dt$.

But I'm pretty confused, what should i do in the 2D case. I sense - i'm not sure -, that now I have a $P \left( x, y \right) \equiv P \left( x, y, t \right)$ probability for the system so I can only calculate $\left< x_{t}^{2} y_{t}^{2} \right>$, but I can cut that into two parts:

$$ \left< x_{t}^{2} y_{t}^{2} \right> = \iint x^{2} \cdot y^{2} P \left( x, y \right)dxdy = \int x^{2} f_{1}(x) dx \cdot \int y^{2} f_{2}(y) dy = \left< x^{2} \right> \cdot \left< y^{2} \right> $$ because we assumed, that jumping into either way is independent from everything. (I actually didn't even know why did i wrote this, just trying to show the actual confusion in my head...) Now I have two 1D rand-walk problem, where jumping into one or to the other way is equally $P = \frac{1}{4}$. But if I do the same calculations I also get $2Dt$ for $\left< x^{2} \right>$ and $\left< y^{2} \right>$ and then get, that $\sqrt{\left< r_{t}^{2} \right>} = \sqrt{\left< x_{t}^{2} \right> + \left< y_{t}^{2} \right>} = \sqrt{2Dt + 2Dt} = 2 \sqrt{Dt}$ which looks totally BS to me, but that's just my opinion.
Where did my thinking go wrong??

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    $\begingroup$ I don't understand why you think $\langle r^2\rangle=4Dt$ isn't legitimate. That is the case for 2D diffusion $\endgroup$ – Aaron Stevens Feb 22 '19 at 22:19
  • $\begingroup$ Ohhhh. Ok, I actually made a big google session about the subject, and i couldn't find anything. The problem is, that I actually used the search term "brownian-motion" instead of "diffusion" for some reasons. Now, encouraged by your comment I tried the same with "diffusion", and found a very nice article, which also confirms, that $\left< r^{2} \right> = 4Dt$. Thank you very much for your confirmation too. Eventually it is nice to know, that I didn't do anything wrong... :D $\endgroup$ – Balázs Pál Feb 23 '19 at 11:07
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Just to make this post clear for anyone, who comes here in the future: Case closed, I actually had the right answer in the post, and Aaron Stevens confirmed it in a reply, thank you.

So for the 2D diffusion, $\left< r^{2} \right> = 4Dt$.

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